Closed-form expressions for Fourier integrals Define the following two Fourier integrals
$$
C(k) = \int_0^{\pi/2} \big( \sin x\big)^{2\alpha}\cos (2kx) dx,\quad
S(k) = \int_0^{\pi/2} \big( \sin x\big)^{2\alpha}\sin (2kx) dx,
$$
for some constant $\alpha>0$ and integer $k$. Using symbolic software like MAPLE, we can get
\begin{align}
C(1) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{1}{2(1+\alpha)},\cr
C(2) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{1-\alpha}{2(1+\alpha)(2+\alpha)},\cr 
C(3) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{(1-\alpha)(2-\alpha)}{2(1+\alpha)(2+\alpha)(3+\alpha)},\cr
C(4) &= -\frac{\sqrt{\pi}\Gamma(\frac{1}{2}+\alpha)}{\Gamma(\alpha)}\frac{(1-\alpha)(2-\alpha)(3-\alpha)}{2(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)},\cr
&\ \ \vdots
\end{align}
It is easy to guess that (seems right, though no proof)
$$
C(k) = -\frac{\alpha\sqrt{\pi}}{2}
\frac{\Gamma(\frac{1}{2}+\alpha)\Gamma(k-\alpha)}{\Gamma(1-\alpha)\Gamma(k+1+\alpha)},\quad k=1,2,\cdots.
$$
But the other integral $S(k)$ seems quite different. Here is the first few ones:
\begin{align}
S(1) &= \frac{1}{1+\alpha},\cr
S(2) &= -\frac{2\alpha}{(1+\alpha)(2+\alpha)},\cr
S(3) &= \frac{3\alpha^3-\alpha+2}{(1+\alpha)(2+\alpha)(3+\alpha)},\cr
S(4) &= -\frac{4{\alpha}^{3}-4{\alpha}^{2}+16\alpha}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)},\cr
S(5) &= \frac{5\,{\alpha}^{4}-10\,{\alpha}^{3}+67\,{\alpha}^{2}-14\,\alpha+24}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)},\cr
S(6) &= -\frac{6\,{\alpha}^{5}-20\,{\alpha}^{4}+202\,{\alpha}^{3}-124\,{\alpha}^{2}+
368\,\alpha
}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)(6+\alpha)},\cr
S(7) &= \frac{7\,{\alpha}^{6}-35\,{\alpha}^{5}+497\,{\alpha}^{4}-601\,{\alpha}^{3}+
2736\,{\alpha}^{2}-444\,\alpha+720
}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)(6+\alpha)(7+\alpha)},\cr
S(8)&=-\frac{8\,{\alpha}^{7}-56\,{\alpha}^{6}+1064\,{\alpha}^{5}-2120\,{\alpha}^{4}
+13712\,{\alpha}^{3}-6464\,{\alpha}^{2}+16896\,\alpha
}{(1+\alpha)(2+\alpha)(3+\alpha)(4+\alpha)(5+\alpha)(6+\alpha)(7+\alpha)(8+\alpha)},\cr
&\ \ \vdots
\end{align} 
Since the numerator of $S(k)$ can not be further factorized (except the obvious factor $\alpha$ when $k$ is even), it is unlikely that $S(k)$ has a single close-form expression as $C(k)$ using Gamma functions. My question is whether there is one with two or more terms?
 A: Here's a sum expression for $S(k)$.
We have
$$
S(k) = \int_0^{\pi/2} \big( \sin x\big)^{2\alpha}\sin (2kx) dx
$$
Integrating by parts gives
$$
S(k) =   \frac{(-1)^{(k+1)}}{2 k}  + \frac{\alpha}{k} \int_0^{\pi/2} \big( \sin x\big)^{(2\alpha -1)}\cos (2kx)\cos (x) dx
$$
We can use the expansion (documented e.g. in the manuscripts of H. W. Gould)
$$
\cos (2kx) =
\sum_{r=0}^k
(-1)^r  \frac{k}{r + k}
{r + k
\choose 
2r
}
4^r \big(\sin x \big)^{2r}
$$
to obtain
$$
S(k) =   \frac{(-1)^{(k+1)}}{2 k}  + \frac{\alpha}{k} \sum_{r=0}^k
(-1)^r  \frac{k}{r + k}
{r + k
\choose 
2r
}
4^r \int_0^{\pi/2} \big( \sin x\big)^{(2(r  +\alpha) -1)}\cos (x) dx
$$
Now the integral can be easily evaluated to give  $\big( \sin x\big)^{(2(r  +\alpha))}/(2(r + \alpha))$ where the definite integral becomes  $1/(2(r + \alpha))$ . Hence
$$
S(k) =   \frac{(-1)^{(k+1)}}{2 k}  +  \sum_{r=0}^k
(-1)^r  \frac{\alpha}{r + k}
{r + k
\choose 
2r
}
 \frac{4^r}{2(r + \alpha)}
$$
or
$$
S(k) =   \frac{1 + (-1)^{(k+1)}}{2 k}  +  \sum_{r=1}^k
(-1)^r  \frac{\alpha}{r + k}
{r + k
\choose 
2r
}
 \frac{4^r}{2(r + \alpha)}
$$
which is the desired expession. Due to the general (not: integer) $\alpha$ this generally cannot be summed any further. Writing this as one fraction, the common denominator is $\prod_{r=1}^{k} (r + \alpha)$ as in the examples in the task description.
$\quad \quad \qquad \Box$
A: These integrals can be expressed on closed form thanks to the hypergeometric function $_2F_1$ 
In case of the integral $C(k)$ , there is a relationship between the hypergeometric function of argument $1$ and the Gamma function which allows a simpler closed form.
In case of the integral $S(k)$ which involves the hypergeometric function of argument $-1$ , as far as I know there is no similar relationship. Nether the less, there is a relationship with the Incomplete Beta function, but in the complex domaine since the argument is $-1$ : Formula above.
Done with the invaluable help of WolframAlpha :

$ _2F_1(-2a,k-a;-a+k+1;-1)=(-1)^k(-1)^a(k-a)\text{B}_{-1}(k-a,2a+1)$
$\text{B}_{-1}(k-a,2a+1) \equiv \text{B}(-1;k-a,2a+1)\quad$
is the Incomplete Beta function with argument $-1$.
http://mathworld.wolfram.com/IncompleteBetaFunction.html
